10 gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40° C . Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearely (L = 80 cal/g)

Question

10 gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40° C . Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearely (L = 80 cal/g) (A) 31° C  (B) 22° C  (C) 19° C  (D) 15° C

Tardigrade Admin · Accepted Answer

Let θ be the temperature when thermal equilibrium has reached. Heat gained by ice to be converted to water at θ° C =m L+m × c ×(T-0)=10 × 80+10 × 1 × T 65 t=2200-800=1400 T=(1400/65) ≈ 21.5° C ≈ 22° C

Q. 10 gm of ice cubes at $0^{\circ} C$ are released in a tumbler (water equivalent 55 g) at $4 0^{\circ} C .$ Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearely (L = 80 cal/g)

$3 1^{\circ} C$

$2 2^{\circ} C$

$1 9^{\circ} C$

$1 5^{\circ} C$

Let $θ$ be the temperature when thermal equilibrium has reached. Heat gained by ice to be converted to water at $θ^{\circ} C$ $= m L + m \times c \times (T - 0) = 10 \times 80 + 10 \times 1 \times T$
$65 t = 2200 - 800 = 1400$
$T = \frac{1400}{65} \approx 21. 5^{\circ} C \approx 2 2^{\circ} C$

Q. 10 gm of ice cubes at 0∘C are released in a tumbler (water equivalent 55 g) at 40∘C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearely (L = 80 cal/g)

31∘C

22∘C

19∘C

15∘C